# [FieldTrip] Magnitude of leadfield in beamformer

Schoffelen, J.M. (Jan Mathijs) janmathijs.schoffelen at donders.ru.nl
Sat Oct 30 15:36:34 CEST 2021

```Hi Jinwen,

The physical units of the lead field are units-of-the-stuff-you-measure-with-your-measurement-device / units-of-dipole-moment.

Thus - for EEG - this will be something like V / Am.

I don’t think I fully understand your question: do you mean with ’the resulting magnitude of lead field was 10^-5’ that the resulting signal was of magnitude 10^-5 (which for EEG would be 10 microvolts)? I would say that this does not sound too bad.

Best wishes,
Jan-Mathijs

On 20 Oct 2021, at 10:14, Jinwen Wei via fieldtrip <fieldtrip at science.ru.nl<mailto:fieldtrip at science.ru.nl>> wrote:

Dear all,

My name is Jinwen and I am conducting forward modelling for EEG source reconstruction following https://www.fieldtriptoolbox.org/workshop/oslo2019/forward_modeling/<https://urldefense.com/v3/__https://www.fieldtriptoolbox.org/workshop/oslo2019/forward_modeling/__;!!HJOPV4FYYWzcc1jazlU!oldn94sFtU6sSZ3RhSg1GJfmLd1KDLSj8B0iy5NPqJS8fTZkip453A0marDA7JmkgmcppbJRrIlKPpA\$>. I have a question about the magnitude of lead field matrix.

According to the tutorial, the leadfield describes what would be seen at the sensor level given that a source was active with a given current (1 Am). In the ending part of the tutorial, a realistic source current of 100 nAm was used to plot the lead field and the resulting magnitude of lead field was 10^-5. That is to say, the original lead field had a magnitude of 10^2 because 10^2 *100*10^-9 = 10^-5. However, how come the original lead field has such a huge magnitude. Actually, when I modelled the lead field using standard BEM, I found the magnitude of the original lead field was quite small, like 10^-5. So I am wondering what magnitude of the original lead field is.

Best regards,
Jinwen
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